+0  
 
+1
46
8
avatar+125 

Find the sum of all positive integral values of n for which  \(\frac{n+6}{n}\)  is an integer.

MobiusLoops  Sep 13, 2018
 #1
avatar+93352 
0

n can be any multiple of 6 so the sum is infinite.  

 

This answer is compeletely  wrong so skip it.

Melody  Sep 13, 2018
edited by Melody  Sep 14, 2018
 #2
avatar
+1

I don't think you're right, what about n=12?

Guest Sep 14, 2018
 #5
avatar+93352 
+1

Yes you are right guest, I screwed up.

Thanks Heureka :)

Melody  Sep 14, 2018
 #3
avatar+20011 
+1

Find the sum of all positive integral values of n for which

\(\mathbf{\large\dfrac{n+6}{n}}\)

is an integer.

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{n+6}{n}} \\\\ &=& \dfrac{n}{n} + \dfrac{6}{n}\\\\ &=& 1 + \dfrac{6}{n}\\ \hline \end{array}\)

 

So \(n\) are the divisors of \(6\).

 

The 4 divisors of 6 are: \(1,2,3,6\)
The sum is \(1+2+3+6 =\mathbf{ 12}\)

 

laugh

heureka  Sep 14, 2018
 #4
avatar
+3

Listfor(n, 1, 10, (n+6) mod n) =(0, 0, 0, 2, 1, 0, 6, 6, 6, 6). Only 4 numbers that give "0" as the remainder satisfy the condition, as heureka found. They are: 1 + 2 + 3 + 6 =12. After 6, all other numbers will give a remainder of 6.

Guest Sep 14, 2018
 #6
avatar+125 
+1

Thanks for the insight. I'm pretty new to integrals.smiley

MobiusLoops  Sep 14, 2018
 #7
avatar+93352 
0

Hi MobiusLoops,

I do not believe you are using the word correctly.

This is an Integer or integer value question.  

 

Integrals are completely different.  They are a part of calculus and nothing to do with integer values.

I understand the confusion, Integral is an normal English word and  integral and integer would have the same origin in speech.  But in maths the two meanings are very different.

Melody  Sep 14, 2018
 #8
avatar+125 
0

Melody, I don't think so. My math question stated that question sooo... I don't know if they made a mistake or I'm not sure.

MobiusLoops  Sep 17, 2018

28 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.